Problem: $f(x, y, z) = (xy, \ln(xz), y^2z^2)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(x + y, 1, 2yz^2 + 2zy^2)$ (Choice B) B $\left(y, \dfrac{1}{x}, 0 \right)$ (Choice C) C $\left(0, \dfrac{1}{z}, 2zy^2 \right)$ (Choice D) D $(x, 0, 2yz^2)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $y$, we'll treat $x$ and $z$ as if they were constants. Therefore, $f_y = (x, 0, 2yz^2)$.